# Covering projection is monomorphism

Let $p:(E,e_0) \rightarrow(X,x_0)$ be a covering projection. Show that $p \sharp: \pi_{1}(E,e_0) \rightarrow \pi_{1}(X,x_0)$ is a monomorphism.

I was wondering here do I need to prove this is a homomorphism?

As I'm confused as I know you have to prove that it is an injection. Which is just by considering what is in the kernel of $\pi_{1}(E,e_0) \rightarrow \pi_{1}(X,x_0)$, something that is homotopic to the constant map, then you just lift the homotopy to the covering space to get something homotopic to the constant map in $(E,e_0)$.

However, is this enough or do you need to show homomorphism? How would you show a homomorphism? would you need to do concatenation of loops?

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In most books on algebraic topology it is proved that any continuous map $f$ induces a homomorphism on $\pi_1$s. It's actually not so hard: if $\gamma_i$ are loops in the domain then $f \circ (\gamma_1 * \gamma_2) = (f \circ \gamma_1) * (f \circ \gamma_2)$. – Dylan Moreland May 21 '12 at 0:49
It is not enough to show that the kernel is trivial. The statement "$\varphi$ is injective $\iff$ $\ker(\varphi) = 0$" requires that $\varphi$ is a homomorphism. For a simple counterexample, the map $x \mapsto x^2$ on $(\mathbb{R}, +)$ has trivial "kernel" but is certainly not injective. – Henry T. Horton May 21 '12 at 0:56
And I suppose you should also prove that the map is well defined first. – Dylan Moreland May 21 '12 at 1:07

If you are not sure about it being a homomorphism, what you should try and prove is that for any map of topological spaces $p:X \to Y$ the induced map $p_*:\pi_1(X,x) \to \pi_1(Y,p(x))$ is a homomorphism. What is this map?
Let $[f]$ be a class in $\pi_1(X,x)$. Then the induced map is $p_*([f]) = [p \circ f]$ where $p \circ f$ is the composite $f:I \to X$ with $p:X \to Y$. Then check that this map is a homomorphism. Recall if we have paths $f:x \to y$ and $g:y \to z$ then $g \cdot f$ is the map obtained by 'going twice as fast' and that this passes through equivalence class via $[g]\cdot[f] = [g \cdot f]$.